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Hilbert spaces

  • Thread starter Raven2816
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  • #1
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{T_a} is an orthonormal system (not necessarily countable) in a Hilbert space H. x is an arbitrary vector in H.


i must show that the inner product <x, T_a> is different fron 0 for at most countably many a.

i'm not even quite sure where to begin. i know that the inner product is the sum(B_aT_a) in a hilbert space, but i'm sure what other information i have to work with.

thanks
 

Answers and Replies

  • #2
HallsofIvy
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Can you sum an uncountable number of non-zero terms?
 
  • #3
Dick
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And what can you say about the sum of |<x,T_a>|^2?
 
  • #4
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could i use the Bessel inequality?
 
  • #5
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or if i fix an e>0...for how many a can it be true that the absolute value of <x, T_a> is greater than e?
 
  • #6
matt grime
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Take the hints you were given. It is impossible for this vector to have a length because you cannot sum an uncountable number of strictly positive terms. Prove it.
 
  • #7
Dick
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or if i fix an e>0...for how many a can it be true that the absolute value of <x, T_a> is greater than e?
This is on the right track. Now answer your own question. How many? Be careful though - how do you know sum |<x,T_a>| converges? Hint: in general, it doesn't.
 
  • #8
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that's a great question. how DO i sum the absolute value of inner products when there are infinite terms? it wont converge. so for a fixed e>0, there are countably many a for which <x, T_a> is greater than e....but why
 
  • #9
Dick
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What does Bessel's inequality tell you? As for how many terms in a convergent infinite series can be greater than some positive number e, can the number even be infinite?
 
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  • #10
HallsofIvy
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that's a great question. how DO i sum the absolute value of inner products when there are infinite terms? it wont converge. so for a fixed e>0, there are countably many a for which <x, T_a> is greater than e....but why
You learned in calculus how to sum an infinite number of terms! But those were all countable sums. Precisely what is meant by "linear combination" here? If there were an uncountable number of terms with non-zero inner product that would give a linear combination of the basis vectors with an uncountable number of non-zero terms. Is that possible the way "linear combination" is defined in a Hilbert space?
 
  • #11
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i know there are countably many nonzero terms...because eventually i will have terms that are no longer linearly independent.
 
  • #12
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i know there are countably many a such that for e>0 the absolute value of <x, T_a> is greater than e. i know that i can't sum an uncountable number of strictly positive terms because it doesn't converge. i just dont understand everything in the middle that should tie this together.
 
  • #13
matt grime
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I don't follow you.

You claim that you can show that if <x,T_a> is non-zero for uncountably many a (which implies <x,x> is infinite), then it is a contradiction. But that was all you were asked to show.
 
  • #14
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i know that <x, T_a> is non-zero for countably many a. i'm just struggling to show that this is in fact true.
 
  • #15
matt grime
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Um. But you just said that if it were false, then you can show a contradiction. Therefore it is true. You can't sum an uncountable number of positive terms. And you know

<x,x>^2 => sum <x,T_a>^2

and the RHS doesn't exist if there are not countably many non-zero terms.
 
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  • #16
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so i can say that since that equality is true, and i can't sum uncountable positive terms, that obviously there are countably many a where the inner product is nonzero? is that a complete answer? it makes intuitive sense, but is it legit?
 
  • #17
matt grime
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Of course it is legit. Why wouldn't it be? You've shown not(B) implies not(A), thus A implies B.
 
  • #18
Dick
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so i can say that since that equality is true, and i can't sum uncountable positive terms, that obviously there are countably many a where the inner product is nonzero? is that a complete answer? it makes intuitive sense, but is it legit?
WHAT equality is true? I haven't seen any equalities playing a big role here. And the reason why you can't have an uncountable number of positive terms in a convergent series isn't because it's 'obvious'. I'm not even sure any of this is 'intuitive'. I think you are going to have to state more of this in the form of a 'proof'.
 
  • #19
matt grime
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I think he meant inequality.
 
  • #20
Dick
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